Question

A school awarded 58 medals for honesty, 20 for punctuality and 25 for obedience. If these medals were bagged by a total of 78 students and only 5 students got all the three medals. Find the number of students who received medals for exactly two of the three medals?

Answer 1

Hint: We start solving the problem by assigning the variables to the number of students obtained each medal. We then use the fact $n\left( T \right)=n\left( H \right)+n\left( P \right)+n\left( O \right)-n\left( H\cap P \right)-n\left( P\cap O \right)-n\left( H\cap O \right)+n\left( H\cap P\cap O \right)$ to find the number of students who got two or more medals. We then use $n\left( H\cap P\cap {{O}^{c}} \right)=n\left( H\cap P \right)-n\left( H\cap P\cap O \right)$, $n\left( P\cap O\cap {{H}^{c}} \right)=n\left( P\cap O \right)-n\left( H\cap P\cap O \right)$ and $n\left( H\cap O\cap {{P}^{c}} \right)=n\left( H\cap O \right)-n\left( H\cap P\cap O \right)$ to get the number of students who received exactly two medals.

Complete step-by-step answer:
According to the problem students are awarded with 58 medals for honesty, 20 for punctuality and 25 for obedience. If these medals were bagged by a total of 78 students and only 5 students got all the three medals. We need to find the number of students who received medals for exactly two of the three medals.
Let us assume the total number of students who got medals be $n\left( T \right)$ and we have $n\left( T \right)=78$.
Let us assume the number of students who got honesty medal be $n\left( H \right)$ and we have $n\left( H \right)=58$.
Let us assume the number of students who got punctuality medal be $n\left( P \right)$ and we have $n\left( P \right)=20$.
Let us assume the number of students who got obedience medal be $n\left( O \right)$ and we have $n\left( O \right)=25$.
We have 5 students who bagged all the three medals. So, we get $n\left( H\cap P\cap O \right)=5$.
We know that $n\left( T \right)=n\left( H \right)+n\left( P \right)+n\left( O \right)-n\left( H\cap P \right)-n\left( P\cap O \right)-n\left( H\cap O \right)+n\left( H\cap P\cap O \right)$.
$\Rightarrow 78=58+20+25-n\left( H\cap P \right)-n\left( P\cap O \right)-n\left( H\cap O \right)+5$.
$\Rightarrow n\left( H\cap P \right)+n\left( P\cap O \right)+n\left( H\cap O \right)=108-78$.
$\Rightarrow n\left( H\cap P \right)+n\left( P\cap O \right)+n\left( H\cap O \right)=30$.
We got number of students who got at least two medals as 30. But we need the total number of students who got exactly two medals.
We know that the number of students who got only honesty and punctuality medals but not obedience medals is $n\left( H\cap P\cap {{O}^{c}} \right)=n\left( H\cap P \right)-n\left( H\cap P\cap O \right)$.
We know that the number of students who got only punctuality and obedience medals but not honesty medals is $n\left( P\cap O\cap {{H}^{c}} \right)=n\left( P\cap O \right)-n\left( H\cap P\cap O \right)$.
We know that the number of students who got only honesty and obedience medals but not punctuality medals is $n\left( H\cap O\cap {{P}^{c}} \right)=n\left( H\cap O \right)-n\left( H\cap P\cap O \right)$.
We need to find the value of $n\left( H\cap P\cap {{O}^{c}} \right)+n\left( P\cap O\cap {{H}^{c}} \right)+n\left( H\cap O\cap {{P}^{c}} \right)$.
So, we have $n\left( H\cap P\cap {{O}^{c}} \right)+n\left( P\cap O\cap {{H}^{c}} \right)+n\left( H\cap O\cap {{P}^{c}} \right)=n\left( H\cap P \right)-n\left( H\cap P\cap O \right)+n\left( P\cap O \right)-n\left( H\cap P\cap O \right)+n\left( H\cap O \right)-n\left( H\cap P\cap O \right)$.
$\Rightarrow n\left( H\cap P\cap {{O}^{c}} \right)+n\left( P\cap O\cap {{H}^{c}} \right)+n\left( H\cap O\cap {{P}^{c}} \right)=n\left( H\cap P \right)+n\left( P\cap O \right)+n\left( H\cap O \right)-3n\left( H\cap P\cap O \right)$.
$\Rightarrow n\left( H\cap P\cap {{O}^{c}} \right)+n\left( P\cap O\cap {{H}^{c}} \right)+n\left( H\cap O\cap {{P}^{c}} \right)=30-\left( 3\times 5 \right)$.
$\Rightarrow n\left( H\cap P\cap {{O}^{c}} \right)+n\left( P\cap O\cap {{H}^{c}} \right)+n\left( H\cap O\cap {{P}^{c}} \right)=30-15$.
$\Rightarrow n\left( H\cap P\cap {{O}^{c}} \right)+n\left( P\cap O\cap {{H}^{c}} \right)+n\left( H\cap O\cap {{P}^{c}} \right)=15$.
So, we have found the number of students who got exactly two medals as 15.
∴ The number of students who got exactly two medals is 15.

Note: We can also solve the problem by drawing venn diagrams using the given information. Drawing venn diagrams gives a better picture for the representation of the information. We can also find the probabilities using the given number of medals as the total number of students who received medals is given. Similarly, we can expect problems to find the conditional probability of students who got three students when at least two students received medals.